Help with finding normal modes of a bar swinging on a string

[BiotFartLaw]

Homework Statement

A bar with mass m and length l is attached at one end to a string (also length l) and is swinging back and forth. Find the normal modes of oscillation.

Homework Equations

L=T-U, and the Lagrange-euler equation, I=(1/12)ml^2

The Attempt at a Solution

So my idea is this. Use the two angles \theta (angle of string from the vertical) and \phi (angle of the bar from the vertical) as the generalized coordinates and set up a Lagrangian to get two (probably coupled) differential equations. THen I'm guessing two different eigenfrequencies will pop out when I solve them.

My problem is that I'm not sure what the (translational) kinetic energy is. (The rotational would just be I*ω^2 and this ω^2 would, in the end, give me the two frequencies [?]) I've tried finding \dot{x} and dot{y} using sines and cosines of the angles, but when I do (and I make appropriate approximations) I'm left with no dot{\theta} or dot{\phi}terms which doesn't seem right.
But I'm also not sure if it's as simple as T=ml^2(dot{\theta}+dot{phi}).

Thanks

 

[vela]

The easiest thing to do is simply write down the x and y coordinates of the center of mass of the rod in terms of the two angles and then calculate $\frac{1}{2}(\dot{x}^2+\dot{y}^2)$.

 

[BiotFartLaw]

Thanks. A new question though:

What is ω (from the rotational KE) in terms of the two angles? Usually \omega = dot{\theta}. But that's only for one angle. Because if ω isn't in terms of the angle(s) then the rotational KE drops out when you take the derivatives for L which doesn't seem right.

Thanks again.

 

[vela]

From the way you defined the angles, you should have $\omega=\dot{\phi}$, right? Changing that angle corresponds to the bar rotating whereas changing $\theta$ only causes translation of the bar.